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Summation of a set = kn?

by Nickclark
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Nickclark
#1
Feb19-13, 07:41 AM
P: 13
i need to find a series of numbers up to n that will add up to kn
x1 + x2 + x3+ ... + n = kn
where k is a constant.

this is part of a long complex problem once this sum is found it will finally be solved.
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Simon Bridge
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Feb19-13, 08:18 AM
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x1 + x2 + x3+ ... + n = kn
Do you mean ##x_1+x_2+x_3+\cdots +x_{n-1}+x_n = nk##?
How about ##x_1=x_2=\cdots =x_{n-1}=x_n = k##?
Nickclark
#3
Feb19-13, 04:45 PM
P: 13
no, i need the last number to be n.
some how i need to find a set of numbers when added together all i get is a multiple of the last number.

Mute
#4
Feb19-13, 05:38 PM
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Summation of a set = kn?

Quote Quote by Nickclark View Post
no, i need the last number to be n.
some how i need to find a set of numbers when added together all i get is a multiple of the last number.
Do you have any restrictions on that set of numbers or the parameters? e.g., how many terms can you have in the sum? Is k any real number, or is it limited to certain values or sets (like the integers or rationals, etc)?
Simon Bridge
#5
Feb19-13, 07:49 PM
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Quote Quote by Nickclark View Post
no, i need the last number to be n.
some how i need to find a set of numbers when added together all i get is a multiple of the last number.
From what you've said, k can be any integer, and there can be any number of terms in the sum ... so you appear to be saying you want to find any set of numbers ##\{x_i\}## so that $$n+\sum_{i=1}^m x_i = kn$$ (the last number is n as specified) .. which suggests that $$(k-1)n=\sum_{i=1}^m x_i$$ and again I can say that ##x_1 = x_2 = \cdots =x_{m-1}=x_m = \frac{n}{m}(k-1)##

... which means that you have to be more specific about your problem if you are to get sensible answers. I'm guessing the x's can't be just any old numbers for example, and k is not supposed to be a specific integer? But I'm guessing - don't make people guess.
Nickclark
#6
Feb20-13, 01:42 PM
P: 13
yeah, i should've added more restrictions.
well the numbers must increase as we go on, for example:
1+2+3+...+n=n(n+1)/2 right, but this is of second order
i want to add numbers that increase up to n, and their sum is of order 1.
Mute
#7
Feb20-13, 08:32 PM
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Quote Quote by Nickclark View Post
yeah, i should've added more restrictions.
well the numbers must increase as we go on, for example:
1+2+3+...+n=n(n+1)/2 right, but this is of second order
i want to add numbers that increase up to n, and their sum is of order 1.
You still need to be a bit more specific. You still haven't told us if k has to be an integer or rational or real or positive or negative, etc. It's also not clear how many terms you have in your sum - it looks like it's n terms where the last term is also equal to n, but you haven't told us that explicitly - we're still just guessing. Please tell us precisely what you want.
Simon Bridge
#8
Feb21-13, 01:46 AM
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If this is a problem that has been supplied to you, then please provide the exact text of the problem and the context.

I'll add that we don't know that the numbers have to be integers of if they can be real, and so on. eg.

##\sum_{i=1}^n = n(n+1)/2 = kn \text{ if } k=(n+1)/2##
But what do you mean "the sum is of order 1"?
You mean that kn cannot be quadratic in n? (k cannot depend on n?)

Maybe you are thinking more like:
x1=1, x2=2, x3=3=n, then 1+2+3=6=2n so k=2
that the idea?
Nickclark
#9
Feb22-13, 10:43 AM
P: 13
Well the problem is not about this sum, if you find a series like the one described the problem will be solved.
i have submitted the following answer:

2+4+8+16+32+64+...+2^k = 2(1+2+4+8+...+2^(k-1) = 2(2^k -1)/2-1 = 2(2^k -1)
but 2^k = n
k = log (base 2) n right, then substitute in the sum:
2(n-1) which is linear that is (of order one).
the idea of the problem is that i want to get to n with the least sum possible, if i went with the quadratic equation that would have taken too much time O(n^2), so this sum will save so much time O(n)
it's an algorithm problem.
Simon Bridge
#10
Feb22-13, 08:09 PM
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So to get good help you will need to describe the problem properly.


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